ZOJ 2319 (二维的LIS)
来源:互联网 发布:冷钢刀怎么样啊 知乎 编辑:程序博客网 时间:2024/06/11 23:14
The most prestigious sports club in one city has exactly N members. Each of its members is strong and beautiful. More precisely, i-th member of this club (members being numbered by the time they entered the club) has strength Si and beauty Bi. Since this is a very prestigious club, its members are very rich and therefore extraordinary people, so they often extremely hate each other. Strictly speaking, i-th member of the club Mr X hates j-th member of the club Mr Y if Si <= Sj and Bi>= Bj or if Si >= Sj and Bi <= Bj (if both properties of Mr X are greater then corresponding properties of Mr Y, he doesn��t even notice him, on the other hand, if both of his properties are less, he respects Mr Y very much).
To celebrate a new 2005 year, the administration of the club is planning to organize a party. However they are afraid that if two people who hate each other would simultaneouly attend the party, after a drink or two they would start a fight. So no two people who hate each other should be invited. On the other hand, to keep the club prestige at the apropriate level, administration wants to invite as many people as possible.
Being the only one among administration who is not afraid of touching a computer, you are to write a program which would find out whom to invite to the party.
This problem contains multiple test cases!
The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.
The output format consists of N output blocks. There is a blank line between output blocks.
Input
The first line of the input file contains integer N - the number of members of the club. (2 <= N <= 100 000). Next N lines contain two numbers each - Si and Birespectively (1 <= Si, Bi <= 109).
Output
On the first line of the output file print the maximum number of the people that can be invited to the party. On the second line output N integers - numbers of members to be invited in arbitrary order. If several solutions exist, output any one.
Sample Input
1
4
1 1
1 2
2 1
2 2
Sample Output
2
1 4
分析: 对于N对(X,Y),找出最长递增子序列。
解法1,DP / LIS,对X从小到大排序,依次添加Y,然后对Y进行LIS,
解法2,保存Y在最外层时的 对X从小到大排序,依次添加Y,用线段树查询小于Y的情况的最大层数。
代码:
#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 100005;int dp[N]; //dp[i]表示最长上升子序列长度为i时子序列末尾的最小B值int mark[N]; //mark[i]表示以第i个people结尾的最长上升子序列的长度struct Member { int s, b; int id; bool operator < (const Member &x) const { if(s == x.s) return x.b < b; return s < x.s; }} a[N];int main() { int n; while(~scanf("%d", &n)) { for(int i = 1; i <= n; i++) { scanf("%d%d", &a[i].s, &a[i].b); a[i].id = i; } sort(a+1, a+n+1); int max_len = 0; dp[++max_len] = a[1].b; mark[1] = max_len; for(int i = 2; i <= n; i++) { if(dp[max_len] < a[i].b) { dp[++max_len] = a[i].b; mark[i] = max_len; } else { int k = lower_bound(dp+1, dp+1+max_len, a[i].b) - dp; dp[k] = a[i].b; mark[i] = k; } } printf("%d\n", max_len); for(int i = n; i >= 1; i--) { if(mark[i] == max_len) { printf("%d", a[i].id); if(max_len > 1) printf(" "); max_len--; } } printf("\n"); } return 0;}
- ZOJ 2319 (二维的LIS)
- ZOJ 2319 Beautiful People(最长单调递增子序列)(二维LIS)
- ZOJ 2319 Beautiful People (LIS的变形)
- ZOJ 2319 Beautiful People(LIS二分法+路径输出)
- 【LIS变形】 zoj 2319 Beautiful People
- acd - 1216 - Beautiful People(二维LIS)
- hdu1069 Monkey and Banana(二维LIS)
- zoj 2319 二维的单调递增子序列
- jdfz-2764 二维LIS
- zoj 11986 LIS 二分
- HDU-1051 Wooden Sticks 【二维LIS(STL应用)+排序】
- ZOJ 3349 Special Subsequence(LIS+线段树优化)
- 20161026的考试】KMP,二维LIS,DAG删一个点求最长路径最小值(BZOJ 3832)
- ZOJ 2319【二维最长递增子序列】
- ZOJ 2859 二维RMQ(模板)
- zoj 2859(二维线段树)
- 二维LIS Gym100820G Racing Gems
- zoj 1108 LIS变形...睡前一水
- Myeclipse调整虚拟内存大小
- 在自己的linux服务器上面部署ShowDoc
- 人见人爱A^B之解题思路
- php面试题
- 既使用maven编译,又使用lib下的Jar包
- ZOJ 2319 (二维的LIS)
- 约瑟夫问题
- 任务管理RHCE-Day5
- 【深入Java虚拟机】之三:类初始化
- C/C++读写BMP文件
- L2-020. 功夫传人
- CodeIgniter首页或尾页没有出现的原因
- 扫描某个包下所有的类,输出所有使用了特定注解的类的注解值
- 解决Dynamic Web Module 3.1 requires Java 1.7 or newer